Question

# $\text{A}$ and $\text{B}$ can do a piece of work in $12$ days , $\text{B}$ and $\text{C}$ can do a piece a work in $15$ days while $\text{A}$ and $\text{C}$ can do a piece of work in $20$ days. At what time $\text{C}$ alone can complete the work ?

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Solution

## $\text{A}$ and $\text{B}$ do a work in$=12$days $\text{A}$ and $\text{B}$ do work in $1$ day $=\frac{1}{12}$$\text{B}$ and $\text{C}$ do a work in $=15$ days $\text{B}$ and $\text{C}$ do a work in $1$ day $=\frac{1}{15}$$\text{A}$ and $\text{C}$ do a work in $=20$ days $\text{A}$ and $\text{C}$ do a work in $1$ day $=\frac{1}{20}$So,$\text{A+B=}\frac{1}{12}\text{→}\overline{)1}$$\text{B+C=}\frac{1}{15}\text{→}\overline{)2}$$\text{A+C=}\frac{1}{20}\text{→}\overline{)3}$By adding all these three equations we get,$\text{A+B+B+C+A+C=}\frac{1}{12}\text{+}\frac{1}{15}\text{+}\frac{1}{20}$$\text{2(A+B+C)=}\frac{1}{12}\text{+}\frac{1}{15}\text{+}\frac{1}{20}\text{=}\frac{1×5+1×4+1×3}{60}\text{=}\frac{5+4+3}{60}\text{=}\frac{12}{60}\text{=}\frac{1}{5}\phantom{\rule{0ex}{0ex}}\left(\text{A+B+C)=}\frac{1}{2×5}\text{=}\frac{1}{10}$ Now $\text{C}$ work alone in one day $=\left(\text{A+B+C)-(A+B)(onedaywork)}\phantom{\rule{0ex}{0ex}}=\frac{1}{10}-\frac{1}{12}=\frac{1×6-1×5}{60}=\frac{6-5}{60}=\frac{1}{60}$From here we say that $\text{C}$ completes the work in $60$ days. Hence $\text{C}$ completes the whole work alone in $60$ days .

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